Identify the function represented by $\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)}$

Question

So first I wrote it out in the terms, and I got

$\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)} = \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...$

I know the power series for $\displaystyle ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ which is similar to the derivative of the power series from above, as

$\displaystyle \frac{d}{dx}(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$

My question is, where do I go from here? How would I make it so the series is similar ln(1+x)? Or am I even going down the right route when it comes to solving this problem? Any help would be appreciated.

Answer 1

Hint A. You have $f'(x)=-\ln(1-x)$. Can you integrate this?

Hint B. Alternatively, use partial fraction decomposition on $\displaystyle\frac{1}{k(k-1)}$.

Answer 2

What you could have done is to differentiate twice to get $$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)''=\sum_{k=2}^\infty x^{k-2}=\frac 1 {1-x}$$making $$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)'=-\log(1-x)$$ $$ \sum_{k=2}^\infty \frac{x^k}{k(k-1)}=x+(1-x) \log (1-x)$$

Answer 3

You're very close. Since $$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots,$$ we have $$- \log (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots.$$ Then integration term by term gives $$- \int \log(1-x) \, dx = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{12} + \frac{x^5}{20} + \cdots.$$ So all you need to do is figure out how to perform the integration. Do be careful, since the RHS should be zero when $x = 0$, so an antiderivative on the LHS should also be zero when $x = 0$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 2}^{\infty}{x^{k} \over k\pars{k - 1}} & = \sum_{k = 2}^{\infty}{x^{k} \over k - 1} - \sum_{k = 2}^{\infty}{x^{k} \over k} = \sum_{k = 1}^{\infty}{x^{k + 1} \over k} - \pars{-x + \sum_{k = 1}^{\infty}{x^{k} \over k}} \\[5mm] & = \pars{1 - x}\pars{-\sum_{k = 1}^{\infty}{x^{k} \over k}} + x \\[5mm] & =\ \bbox[10px,#ffd,border:1px groove navy]{\large\pars{1 - x} \ln\pars{1 - x} + x} \\ & \end{align}